Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON):

Solution:
N=int(raw_input())
P=[float(x) for x in raw_input().split()]
A=[[0 for i in range(N+1)] for j in range(N+1)]
A[1][0]=1-P[0]
A[1][1]=P[0]
for j in range(2,N+1):
    A[j][0]=A[j-1][0]*(1-P[j-1])
    for k in range(1,j+1):
        A[j][k]=A[j-1][k]*(1-P[j-1]) + A[j-1][k-1]*(P[j-1])
print(sum(A[N][N/2+1:]))